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I have run into the following expectation and am not sure what it is.

$ X \sim N(0, \sigma^2)$

Then $\mathrm{e}^{x}$ is lognormal, but what is the expectation of $\mathbb{E}\left[\exp(X^2)\right]$? $X^2$ is a chi squared distribution, but then what is the distribution of $\exp(X^2)$?

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1 Answer 1

Write out the definition of the expectation: $$ \mathbb{E}\left(\exp(X^2)\right) = \int_{-\infty}^{\infty} \mathrm{e}^{x^2} \frac{1}{\sqrt{2\pi} \sigma} \exp\left(-\frac{1}{2} \frac{x^2}{\sigma^2} \right) \mathrm{d} x $$ When does this integral converge?

As to the distribution law of $Y=\exp(X^2)$, assuming $y>1$ $$ F_{Y}(y) = \mathbb{P}\left(\exp(X^2) < y \right) = \mathbb{P}\left(X^2 < \log(y)\right) = F_{X^2}(\log(y)) $$ Therefore the probability density function of $Y$: $$ f_Y(y) = F_{Y}^\prime(y) = \frac{1}{y} f_{X^2}(\log(y)) = \frac{1}{\sqrt{2 \pi } \sigma ^2 \sqrt{\log (y)}} y^{-\frac{1}{2 \sigma ^4}-1} $$

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